Question

A long straight wire carries a current $i$. a particle having a positive charge $q$ and mass $m$, kept at distance $x_0$ from the wire is projected towards it with speed $v$. If the closest distance of approach of charged particle to wire is $x_{min}=x_0{e}^{-Y\pi mv/{\mu }_{0}qi}$. Find Y.

Answer 1

Magnetic field due to long wire is: $\overrightarrow{B}=-\frac{{\mu }_{0}i\hat{k}}{2\pi x}$
Let velocity of the charged particle at any instant be: $\overrightarrow{v}=(v_x\hat{i}+v_y\hat{j})$

Magnetic force on the charge, $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})=-\frac{{\mu }_{0}qi{v}_y}{2\pi x}\hat{i}+\frac{{\mu }_{0}q}{2\pi x}\hat{j}......\left(i\right)$
From Eq. $\left(i\right)$, $y$ component of acceleration is:
$a_y=\frac{d{v}_y}{dt}=\frac{{\mu }_{0}qi{v}_x}{2\pi mx}=\frac{{\mu }_{0}q}{2\pi mx}i\frac{dx}{dt}$

At minimum separation, velocity of particle is $-v\hat{j}$
(closest distance will be when velocity becomes parallel to wire or in -ve y-direction)
${\int }_0^{-v}d{v}_y=\frac{{\mu }_{0}qi}{2\pi m}{\int }_{x_0}^{x_{min}}\frac{dx}{x}\Rightarrow -v=\frac{{\mu }_{0}qi}{2\pi m}\ln \left|\frac{x_{min}}{x_0}\right|$

Thus, $x_{min}=x_0{e}^{-2\pi mv/{\mu }_{0}qi}$